A new way to represent functions as series

Theorem 3.1

Let c₁ and c₂ defined as above. Then, ∀h ≥ 3, ∃ c_h ∈ (c₁, c_h–1) such that

If h = 3, we obtain (3.6).

Proof

The proof is by induction. We know that for h = 3 the formula holds true. Now, suppose that it holds true for a number h. We have to prove that it also holds for h + 1. Then, by induction, we can conclude that the formula holds true ∀h ≥ 3. So, suppose that

Consider now this formula for h + 1:

We know that ∀h ≥ 2

This can be written as

and it implies that

Hence, we have that

Since the first expression equals f(3), also the second is equal to f(3), and therefore, we proved the theorem.□

Theorem 3.2

Let f ∈ C^∞([x₀–1, x₀+1]). Let c₁ ∈ (x₀–1, x₀), c₂ ∈ (x₀, x₀+1) such that f(x₀)–f(x₀–1) = f′(c₁), f(x₀ + 1)–f(x₀) = f′(c₂). Then, ∀h ≥ 3, ∃ c_h ∈ (c₁, c_{h – 1}) such that

Proof

Analogous to the one of the previous Theorem.□

Definition 4.1

Let c_j be the points defined before. For j ≥ 2 we will let

Proposition 4.1

Let f be a function with the properties in Theorem 3.2. Then f is expandable if and only if f^{(h – 1)}(c_h)n(x₀, h – 1) → 0 when h → +∞.

Proof

From Theorem 3.2, we know that

If f is expandable, since for h → +∞ we have

in order to obtain (4.1), we must have f^{(h – 1)}(c_h)n(x₀, h – 1) → 0.□

Proposition 4.2

∀σ ≥ 2, σ ∈ ℕ,

Proof

First of all, notice that, since f(x₀) – f(x₀ – 1) = f′(c₁) and f(x₀ + 1) – f(x₀) = f′(c₂), we have that

Furthermore,

Hence

Since

we can write

From the formulae that give c_j, we know that f″(c₃)–f″(c₁) = f″′(c₄)l(x₀, 3). Therefore,

Iterating this, we obtain the above theorem.□

Lemma 5.1

If l(x₀, i₀) ≤ 1 for a certain i₀ ∈ ℕ, i₀ ≥ 2, then l(x₀, i₀) ≤ 1 ∀ i ≥ i₀.

Proof

First of all, notice that l(x₀, i) > 0 because c_i > c₁ (i > 1). Furthermore: l(x₀, i + 1) < l(x₀, i) ∀i ≥ 2, since c_i+1 < c_i. Generalizing this, we can say that c_i+j < c_i, ∀i ≥ 2, ∀j ≥ 1. So, if l(x₀, i₀) ≤ 1, this inequality holds true ∀i ≥ i₀.□

Theorem 5.1

Let f ∈ C^∞([x₀ – 1, x₀ + 1]). Then, if there exists an index i₀ ≥ 2 such that l(x₀, i₀) ≤ 1 and, ∀x ∈ (x₀ – 1, x₀ + 1), |f⁽ⁱ⁾(x)| → 0 when i → +∞, the function is expandable.

Proof

We have to show that f^{(h – 1)}(c_h)n(x₀, h – 1) → 0 when h → +∞. This is equivalent to prove that |f^{(h – 1)}(c_h)n(x₀, h – 1)| → 0.

Since |f⁽ⁱ⁾(x)| → 0 ∀x ∈ (x₀ – 1, x₀ + 1), and c_h ∈ (x₀ – 1, x₀ + 1), we obtain |f^{(h – 1)}(c_h)| → 0; we have to show that n(x₀, h – 1) ↛ ∞, so that the product goes to 0.

Since l(x₀, i₀) ≤ 1, by Lemma 5.1 l(x₀, i) ≤ 1 ∀i ≥ i₀. Hence,

because it is constant. If i₀ = 2, consider ∏b=2i0−1 l(x₀, b) = 1.□

Theorem 5.2

Let f ∈ C^∞([x₀ – 1, x₀ + 1]). If |f⁽ⁱ⁾(x)| ≤ M ∈ ℝ₊ ∀x ∈ (x₀ – 1, x₀ + 1) and ∀i ∈ ℕ, and if n(x₀, i) → 0 when i → +∞, then f is expandable.

Proof

We have to show that f^{(h – 1)}(c_h)n(x₀, h – 1) → 0 when h → +∞. Since |f^{(h – 1)}(c_h)n(x₀, h – 1)| ≤ M n(x₀, h – 1) by hypothesis, and n(x₀, i) → 0, the product goes to 0 and we have finished.□

Theorem 5.3

∑n=1+∞ log(a_n) = –∞ ⇔ ∏n=1+∞ a_n = 0, where a_n > 0 ∀n > 0.

Proof

For details, see [3].

If the sum is –∞, log( ∏n=1+∞ a_n) → –∞ and this happens when the argument ∏n=1+∞ a_n → 0. Vice versa, if ∏n=1+∞ a_n → 0, we have that log( ∏n=1+∞ a_n) → –∞ and so the sum tends to –∞.□

Theorem 5.4

n(x₀, i) → 0 when i → +∞ ⇔ ∑n=2+∞ log(l(x₀, n)) = –∞.

Proof

Let a_n = l(x₀, n + 1). Applying Theorem 5.3, we have that

if and only if

Writing

we have proved the theorem.□

Theorem 5.5

If there exists n₀ ≥ 2 such that l(x₀, n₀) < 1e , then n(x₀, i) → 0 when i → +∞.

Proof

If l(x₀, n₀) < 1e , l(x₀, n) < 1e ∀n ≥ n₀ (for the same reason why Lemma 5.1 holds true). Hence, log(l(x₀, n)) < log( 1e ) = –1. Thus,

Therefore, ∑n=2+∞ log(l(x₀, n)) = –∞. Applying Theorem 5.4, we conclude.□

Theorem 5.6

Let f ∈ C^∞([x₀ – 1, x₀ + 1]) such that ∀x ∈ (x₀ – 1, x₀ + 1) |f⁽ⁱ⁾(x)| ≤ M ∈ ℝ₊. Then, if there exists a j ≥ 2 such that l(x₀, j) < 1e , f is expandable.

Proof

We have to prove that f^{(h – 1)}(c_h)n(x₀, h – 1) → 0. We have

because under the said hypothesis n(x₀, h – 1) → 0 by Theorem 5.5.□

Example 6.1

Let f(x) = e^ax, a ∈ ℝ, a ∈ (–1, 1), a ≠ 0. |f⁽ⁱ⁾(x)| = |a|ⁱ e^ax → 0. We want to find some of the c_h. To do this, we have to solve

As noticed before, l(x₀, 2) = 1 and |f⁽ⁱ⁾(x)| → 0 ∀a ∈ (–1, 1), a ≠ 0, so by Theorem 5.1, f is expandable for these values of a. To write its series, notice that

and so

Hence, we have that

which implies

and this in turn yields

since ae^ac₁ = e^a(x₀–1)(e^a – 1). This becomes

Writing some terms, we have

Now, notice that when a = ±1, |f⁽ⁱ⁾(x)| = e^±x ≤ M ∈ ℝ₊ since it is independent of i. We wonder, whether the function is expandable also for these values, or not. We know that the derivatives are limited, so we just have to verify that ∃j : l(x₀, j) < 1e in both cases. Take, for instance, j = 4; we have

Since both these values are < 1e , we can expand f. We get a new interesting representation of e, i.e.

□

Example 6.2

Let f(x) = sin(ωx), ω ∈ (0, 1). In this interval, we have that |f⁽ⁱ⁾(x)| ≤ |ω|ⁱ → 0 when i → +∞. To know when f is expandable and to write its series, we evaluate some points c_j; notice that we can choose any x₀; here, for the sake of simplicity, we will just consider the particular case x₀ = 1. We have

where p₁, p₂ ∈ ℤ since cos y = t for 0 < t ≤ 1 yields y = 2p π ± arccos t for some p ∈ ℤ, and c₁ ∈ (0, 1), c₂ ∈ (1, 2) because of Lagrange’s Theorem. So we want to find p₁, p₂ ∈ ℤ such that c₁ ∈ (0, 1) and c₂ ∈ (1, 2). We claim that they are both equal to 0, and that we have to choose the sign +. To prove this, we have to verify that

First of all, notice that the function h(ω) := sin ω – ω cos ω is strictly increasing in (0, 1), and h(0) = 0; therefore, sin ω – ω cos ω > 0 in (0, 1) and sin⁡ωω > cos ω. Since arccos is a decreasing function in its domain of definition, we have arccos( sin⁡ωω ) < ω. Noting also that arccos y = 0 only when y = 1, and that sin⁡ωω ≠ 1 when ω ∈ (0, 1), we can conclude that 0 < 1ω arccos( sin⁡ωω ) < 1. In a similar way we can prove the other inequality.

We can actually verify that when ω ∈ (0, 1)

and

Indeed, both g₁(ω) := 1ω arccos( sin⁡ωω ) and g₂(ω) := 1ω arccos (sin⁡(2ω)−sin⁡ωω) are decreasing functions in (0, 1), and therefore,

Since sup_(0,1) c₂ – inf_(0,1) c₁ ≈ 0.9567289… < 1, we certainly have l(1, 2) = c₂ – c₁ < 1 for ω ∈ (0, 1), and we conclude that ∀ω ∈ (0, 1) we can expand f.

By (4.1), we can write the series

We can write some terms, namely

We now wonder, whether or not f is expandable for ω = 1. To answer this question, we have to find a l(1, j) < 1e . We can easily verify that, for this value of ω, l(1, 4) ≈ 0.21119234 < 1e , so we can expand f ∀ω ∈ (0, 1].

If we want to expand f for an argument in [–1, 0), we can just remark that

since sin ω = –sin(–ω) for any ω.

Furthermore, the following fact is also interesting: consider ω=π4∈(0,1]; we have sin⁡π4=12. Hence, we can write

After some algebraic manipulations, we eventually obtain

□

Example 6.3

Let f(x) = e^{sin x}. We could expand f as usual, but we would not have the exact values of the c_j. In this case it can be better to consider a = sin(x) and expand first e^a. Since, by Example 6.1, ∀a ∈ [–1, 1], a ≠ 0,

we just have to put a = sin x ∈ [–1, 1], sin x ≠ 0. Since sin x ∈ [–1, 1] holds ∀x ∈ ℝ, we just have to exclude those values of x for which sin x = 0. This means that we can expand f ∀x ≠ tπ, t ∈ ℤ. The expansion is obtained by putting sin x instead of a in the expansion of e^a.

We could also expand all the terms with sin x in this equation with the series of Example 6.2, if we wanted.□